Timeline for Better estimator of expected sum than mean
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| Oct 14, 2017 at 19:25 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 13, 2017 at 2:30 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
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| May 3, 2017 at 17:15 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Mar 30, 2017 at 23:37 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 22, 2017 at 21:28 | comment | added | Keith | The best way to think of it is in terms of AB testing. The A and B for the test would be Y and Z. Given Y and Z you want to choose a data source or underlying pdf to sample from. That new sample would be X. Once you have made your decision based on Y and Z you get the X data set sampled from the pdf of your choice. You want to maximize the sum of X. | |
| Feb 22, 2017 at 6:50 | comment | added | user83346 | You mention X in the first paragraph and in the second one you come up with Y and Z, how are the latter two related to the sum of the X? | |
| Feb 21, 2017 at 23:48 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 18, 2017 at 12:37 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 16, 2016 at 16:27 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 15, 2016 at 23:48 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 9, 2016 at 17:21 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 7, 2016 at 0:34 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 7, 2016 at 10:27 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 1, 2016 at 9:22 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 26, 2016 at 16:56 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 21, 2016 at 7:56 | comment | added | Keith | Ugg I swear you know what I mean... Over many trials you are most likely to be correct. | |
| Feb 20, 2016 at 7:22 | comment | added | Glen_b | "where the best bet is the best" doesn't define anything, since you don't define what makes a "best bet" best. It just shifts the "what does 'best' mean" problem around (dIfferent bets will be better as you change your utility function/loss function) | |
| Jan 15, 2016 at 20:39 | answer | added | Keith | timeline score: 1 | |
| Jul 18, 2014 at 21:07 | history | tweeted | twitter.com/#!/StackStats/status/490241559766786048 | ||
| Jul 17, 2014 at 7:26 | comment | added | Keith | So far all I have proven is that it is the same as in the case where you sample from a discrete set and that I can take that set as Bayesian prior information. I have then asked the simplified question here to keep the ball rolling. stats.stackexchange.com/questions/107848/… | |
| Jul 17, 2014 at 3:02 | comment | added | Jessica Collins | @Keith, feel free to answer your own question. I'd be interested in what you've learned. | |
| Jul 13, 2014 at 0:31 | comment | added | Keith | Yep pretty much. The historical purchase data is for two sets of customers exposed to different A/B test variants $Y$ and $Z$ (eg changes in marketing). I want to know which variant is better based on the data sets and the prior information I gave in the question. One of the variants will be chosen as a standard for the future and clearly I want to maximize future revenue. I am open to time series analysis but I don't see how it would add anything. It is close enough to a multi-armed bandit problem that I am making some progress with a variation of Thompson sampling. | |
| Jul 12, 2014 at 20:49 | comment | added | Jessica Collins | So you have historical purchase data of N customers. You want to select which customers are the most profitable? To what end? Are you trying to direct marketing efforts? This doesn't appear to be a multi-armed bandit problem. $Y$ and $Z$ are not stationary distributions you're sampling from. In fact $Y$ and $Z$ are time-series of purchasing events. They could be summed over weekly periods then traditional time-series forecasting methods could project into the future. You could attempt to measure the impact of marketing efforts. Is this close? | |
| Jul 12, 2014 at 17:44 | comment | added | Keith | @JacobMick I am not in BI but after some research I think I can translate for the example I give in the question. A split test is an A/B test and in this case the two variants are called $Y$ and $Z$. They could be represented as two tables with one column each where the rows represent each customer and the values are the dollars spent. The question is which estimator (eg $Y.mean()$) is best to decide which group ($Y$ or $Z$) will produce highest future profit. This would be represented as $X.sum()$ where the data $X$ is the spend for each customer collected after the decision is made. Clear? | |
| Jul 12, 2014 at 13:25 | comment | added | Jessica Collins | Your vocabulary here is a bit foreign to me. Could you edit your original question to include data dictionary (what do the rows and columns in your design matrix represent), along with what you want a model to do? | |
| Jul 12, 2014 at 10:52 | history | edited | Keith | CC BY-SA 3.0 |
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| Jul 12, 2014 at 10:51 | comment | added | Keith | There is another modification to the multi-armed bandit problem which makes the customer revenue example more fitting. A customer can purchase more than one item but a slot machine has one payoff per pull. Also, the fact that you know the previous million results and have to choose and stick with one machine based on that information makes it seem to be a different problem to me since in the multi-armed bandit problem you can change your choice. | |
| Jul 12, 2014 at 10:19 | history | edited | Keith | CC BY-SA 3.0 |
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| Jul 12, 2014 at 8:57 | comment | added | Keith | Yes, I think this is the answer except there is only one round and the reward is $\Sigma X_i$ where the elements are drawn from your choice. The information about each choice is given above. They are exponentially decreasing, most elements are 0 and pdfs are discrete in an complex way. If you post your comment as an answer and include some mention of the specifics in this particular context I will choose it as the answer. | |
| Jul 12, 2014 at 8:44 | history | edited | Keith | CC BY-SA 3.0 |
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| Jul 11, 2014 at 23:37 | comment | added | Jessica Collins | If I'm understanding you correctly, this is analogous to having several choices to sample from. At each round you select a choice to sample from. The choice you selected will yield a reward. You want to maximize this reward. Each choice has an unknown distribution and parameters. This is the multi-armed bandit problem, which is essentially solved. If you have side information about each choice, this is the contextual bandit problem. | |
| Jul 11, 2014 at 23:22 | comment | added | Keith | Sorry, there is a large jargon gap here between differing fields. I still think the revenue example adds clarity but I have added more to help clarify my question. | |
| Jul 11, 2014 at 23:18 | history | edited | Keith | CC BY-SA 3.0 |
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| Jul 11, 2014 at 15:25 | comment | added | whuber♦ | You lost me in the first sentence with "maximal expected." If the $X_i$ are a sample then its mean has an expectation which is a number: what is "maximal" about that? In the following sentences you jump right in with references to "revenue" and "customers" without providing the context to understand how they might be related to the $X_i$. I'm afraid I stopped reading right there and suspect many others who otherwise might be able to help you would be similarly disinterested. Do you think you could edit this post to help people understand what you are doing? | |
| Jul 11, 2014 at 13:53 | history | asked | Keith | CC BY-SA 3.0 |